A sphere and a cube made of the same material | vedclass

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(A) The rate of cooling $R = \frac{dT}{dt} = \frac{e \sigma A (T^4 - T_0^4)}{ms}$.Since $m = \rho V$,we have $R = \frac{e \sigma A (T^4 - T_0^4)}{\rho

Vedclass · Accepted Answer

$\sqrt{\frac{\pi}{6}} : 1$

Vedclass · Answer

(A) The rate of cooling $R = \frac{dT}{dt} = \frac{e \sigma A (T^4 - T_0^4)}{ms}$.Since $m = \rho V$,we have $R = \frac{e \sigma A (T^4 - T_0^4)}{\rho V s}$.For the same material,$e, \sigma, \rho, s$ are constant. Given the same surface area $A$ and temperature $T$,$R \propto \frac{1}{V}$.For a sphere of radius $r$,$A = 4\pi r^2 \Rightarrow r = \sqrt{\frac{A}{4\pi}}$. Volume $V_s = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (\frac{A}{4\pi})^{3/2} = \frac{A^{3/2}}{6\sqrt{\pi}}$.For a cube of side $a$,$A = 6a^2 \Rightarrow a = \sqrt{\frac{A}{6}}$. Volume $V_c = a^3 = (\frac{A}{6})^{3/2} = \frac{A^{3/2}}{6\sqrt{6}}$.Ratio $\frac{R_s}{R_c} = \frac{V_c}{V_s} = \frac{A^{3/2} / 6\sqrt{6}}{A^{3/2} / 6\sqrt{\pi}} = \frac{\sqrt{\pi}}{\sqrt{6}} = \sqrt{\frac{\pi}{6}}$.Thus,the ratio is $\sqrt{\frac{\pi}{6}} : 1$.

$A$ sphere and a cube made of the same material and having the same surface area are heated to the same temperature and kept in the same surroundings. The ratio of their initial rates of cooling will be

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